Answer
$ 7.07 mi$
Work Step by Step
From above figure
$AB = AB' = 135 mi$
The plane is off by distance of BB'
Which can be calculated from isosceles triangle ABB'
BB' is bisected by perpendicular from A
Also $\angle A$ is bisected by perpendicular from A
So, $BB' = 2 AB' sin(1.5°) = 7.07 mi$
